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Proof step 12 from proof refin-del (previous step)

r ∈ st -- r1 ⊦ r ∈ st, r = r1

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ r ∈ st -- r ∧ (∀ r0, sk. r ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r) ∧ st[r0 -- sk] = (st -- r)[r0 -- sk])
Proof step 11:
¬ r1 ∈ st -- r1 ∧ (∀ r0, sk. r1 ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r1) ∧ st[r0 -- sk] = (st -- r1)[r0 -- sk]), r ∈ st -- r1 ⊦ r ∈ st, r = r1

Rule argument

[LEMMAARG : deleteref, [r, st] → [r1, st]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r1, st)]]